Will These Integrals Converge or Diverge as x Approaches Infinity?

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Homework Help Overview

The discussion revolves around predicting the convergence or divergence of integrals as x approaches infinity, specifically focusing on two problems involving rational functions. The original poster references a box of known behaviors for certain integrals to guide their reasoning.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore comparisons between the given integrals and known divergent integrals, questioning the validity of using a diverging integral to infer properties about a smaller integral.
  • Some participants suggest examining the behavior of the integrands as x approaches infinity and comparing them to simpler functions.
  • There is a discussion about the implications of the integrand approaching a constant value and what that might indicate about convergence.

Discussion Status

The discussion is active, with participants providing insights and clarifications. Some guidance has been offered regarding the comparison of integrals, and there is an acknowledgment of a misunderstanding that has been addressed. However, there is still some uncertainty regarding the implications of certain comparisons.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the types of comparisons and conclusions they can draw. There is also a mention of a correction regarding the first problem, indicating that the original poster is still in the process of clarifying their understanding.

Jacobpm64
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Homework Statement


Use the box and the behavior of rational and exponential functions as [tex]x \rightarrow \infty[/tex] to predict whether the integrals converge or diverge.

Here is the box:
[tex]\int^\infty_1 \frac{1}{x^p} dx[/tex] converges for p > 1 and diverges for p < 1.

[tex]\int^1_0 \frac{1}{x^p} dx[/tex] converges for p < 1 and diverges for p > 1.

[tex]\int^\infty_0 e^{-ax} dx[/tex] converges for a > 0.

Problem 1:
[tex]\int^\infty_1 \frac{x^2}{x^4 + 1} dx[/tex]

Problem 2:
[tex]\int^\infty_1 \frac{x^2 - 6x + 1}{x^2 + 4} dx[/tex]

Homework Equations


The ones in the box above.

The Attempt at a Solution


Problem 1:
I know that this integral is less than [tex]\int^\infty_1 \frac{1}{x} dx[/tex]. I also know that [tex]\int^\infty_1 \frac{1}{x} dx[/tex] diverges. This does not help me though because I can not use a diverging integral to say that a smaller integral is also diverging. This is where I'm confused.

Problem 2:
I know that this integral is less than [tex]\int^\infty_1 1 dx[/tex] I also know that [tex]\int^\infty_1 1 dx[/tex] diverges. This does not help me though because I can not use a diverging integral to say that a smaller integral is also diverging. So, I'm confused at the same place on this problem.
 
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For problem 1, isn't (x^2)/((x^4)+1) always less than 1/(x^2) for values of x > 1?

For problem 2, what does the integrand approach as x approaches infinity?
 
[tex]\frac{x^2}{x^4+1}< \frac{x^2}{x^4}= \frac{1}{x^2}[/tex]

For the second one, divide first:
[tex]\frac{x^2+ 6x+ 1}{x^2+ 4}= 1+ \frac{6x- 3}{x^2+ 4}[/itex]<br /> What does that first "1" tell you?[/tex]
 
Oops, I copied down the wrong first problem.

The second one was correct though.

I understand the first one, thanks for that.

I still don't understand what you mean by what does the first "1" tell me.

Wait, does that mean.. that my integral is bigger than the integral of 1, so if the integral of 1 diverges, I can say that my integral also diverges!

woot!

Thanks!

I'm going to post the correct problem for #1, in a new post. Thanks.
 

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